Find an algebraic expression that is equivalent to the expression. $$\sec ( \arctan 4 x )$$
A) $\sqrt { 16 x ^ { 2 } + 1 }$
B) $\sqrt { 1 - 16 x ^ { 2 } }$
C) $\frac { \sqrt { 16 x ^ { 2 } - 1 } } { 16 x ^ { 2 } - 1 }$
D) $\sqrt { 16 x ^ { 2 } - 1 }$
E) $\frac { \sqrt { 16 x ^ { 2 } + 1 } } { 16 x ^ { 2 } + 1 }$

Question

Find an algebraic expression that is equivalent to the expression.​ $$\sec ( \arctan 4 x )$$ ​
A) $\sqrt { 16 x ^ { 2 } + 1 }$
B) $\sqrt { 1 - 16 x ^ { 2 } }$
C) $\frac { \sqrt { 16 x ^ { 2 } - 1 } } { 16 x ^ { 2 } - 1 }$
D) $\sqrt { 16 x ^ { 2 } - 1 }$
E) $\frac { \sqrt { 16 x ^ { 2 } + 1 } } { 16 x ^ { 2 } + 1 }$

Quizplus · Accepted Answer

To find an expression equivalent to $\sec(\arctan(4x))$, consider a right triangle where $\tan(\theta) = 4x$. This means the opposite side over the adjacent side is $4x$. If we let the opposite side be $4x$ and the adjacent side be $1$, the hypotenuse, using the Pythagorean theorem, is $\sqrt{(4x)^2 + 1^2} = \sqrt{16x^2 + 1}$. Since $\sec(\theta) = \frac{1}{\cos(\theta)}$ and $\cos(\theta)$ is the adjacent side over the hypotenuse in a right triangle, $\sec(\theta) = \frac{\text{hypotenuse}}{\text{adjacent}}$, which in this case is $\sqrt{16x^2 + 1}$.

Find an Algebraic Expression That Is Equivalent to the Expression sec⁡(arctan⁡4x)\sec ( \arctan 4 x )sec(arctan4x)

Find an Algebraic Expression That Is Equivalent to the Expression $\sec ( \arctan 4 x )$